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What is the answer to Let Sn=1-2+3-4+...+(-1)n-1 n then S47+S33+S100 equals?

I need an explanation on how to do it also.
For Sn the n should be in the bottom
N-1 should be at the top
In S47 the 47 should be in the bottom In S53 the 53 should be at the bottom. And for S100 the 100 should be in the bottom.

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firstly,we can list some specific numbers for "n"
for example,we choose 1,2,3,4,5,6 to replace "n"
so
S1=1
S2=1-2= -1
S3=1-2+3=2
S4=1-2+3-4= -2
S5=1-2+3-4+5=3
S6=1-2+3-4+5-6= -3
ok,so next,we need to analyse these procedures.
obviously, we can find that
the results of all the odd items are positive and they are 1,2,3...
the results of all the even items are negative and they are -1,-2,-3...
so we can summary the Sn formula?
1.if n is odd , then Sn=(n+1)/2
2.if n is even, then Sn= -n/2
so now it‘s easy to solve S47+S33+S100
S47+S33+S100=(47+1)/2+(33+1)/2+ -100/2= -9
the final answer is - 9

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4?

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